# Find the specific solution of the differential equation dy/dx equals the quotient of 2 times y and x squared with condition

2 answers

###### Question:

Find the specific solution of the differential equation dy/dx equals the quotient of 2 times y and x squared with condition y(-2) = e. (4 points)

a. y equals negative 1 minus 2 divided by x

b. y equals e raised to the negative 2 over x power

c. y equals negative 1 times e raised to the 1 over x power

d. none of these

a. y equals negative 1 minus 2 divided by x

b. y equals e raised to the negative 2 over x power

c. y equals negative 1 times e raised to the 1 over x power

d. none of these

## Answers

B. y = e^(-2/x).

Step-by-step explanation:

dy/dx = 2y / x^2

Separate the variables:

x^2 dy = 2y dx

1/2 * dy/y = dx/x^2

1/2 ln y = = -1/x + C

ln y = -2/x + C

y = Ae^(-2/x) is the general solution ( where A is a constant).

Plug in the given conditions:

e = A e^(-2/-2)

e = A * e

A = 1

So the specific solution is y = e^(-2/x).

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4y}{x^2}\implies\dfrac{\mathrm dy}y=\dfrac4{x^2}\,\mathrm dx[/tex]

Integrate both sides to get

[tex]\ln|y|=-\dfrac4x+C[/tex]

Given that [tex]y(-4)=e[/tex], we have

[tex]\ln|e|=-\dfrac4{-4}+C\implies C=0[/tex]

so the particular solution is

[tex]\ln|y|=-\dfrac4x\iff y=e^{-4/x}[/tex]

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